Trigonometry (11th Edition) Clone

$$(\sin\theta-\cos\theta)(\csc\theta+\sec\theta)=\tan\theta-\cot\theta$$
$$A=(\sin\theta-\cos\theta)(\csc\theta+\sec\theta)$$ - Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}$$ $$\csc\theta=\frac{1}{\sin\theta}$$ Replace into $A$: $$A=(\sin\theta-\cos\theta)(\frac{1}{\sin\theta}+\frac{1}{\cos\theta})$$ $$A=(\sin\theta-\cos\theta)\Big(\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\Big)$$ $$A=\frac{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}{\sin\theta\cos\theta}$$ $$A=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta\cos\theta}$$ (for $(A-B)(A+B)=A^2-B^2$) Now we can separate the numerator to eliminate the denominator: $$A=\frac{\sin^2\theta}{\sin\theta\cos\theta}-\frac{\cos^2\theta}{\sin\theta\cos\theta}$$ $$A=\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}$$ - Quotient Identities: $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Therefore, $$A=\tan\theta-\cot\theta$$